$u, u^2,u^3$

Relevant wiki: Increasing / Decreasing Functions

Consider the RHS as a function of $u$ . Since each of these terms are positive increasing functions, hence their product is a positive increasing function.

This tells us that, for any constant, there is a unique value of $u$ that satisfies it. It is also clear that the range of the RHS is just $(0, \infty )$ . Since the LHS falls into this range, hence there is 1 value of $u$ that satisfies the equation.

Note: As it turns out, $u = 1$ works! Who knew?

Shouldn't this be an equation instead of an inequality? Anyway...

$28 \times 3 \times 2009 = 28^u \times 3^{u^2} \times 2009^{u^3}$

It's pretty obvious that $u = 1$ satisfies this equation.

But, is there any other solution besides this? We need to check.

To do this, we can take the $\log$ of both sides of the equation. This gives:

$\log_{10}(28 \times 3 \times 2009) = \log_{10}(28^u \times 3^{u^2} \times 2009^{u^3})\\ \log_{10} 28^u + \log_{10} 3^{u^2} + \log_{10} 2009^{u^3} = \log_{10}(28 \times 3 \times 2009)\\ (\log_{10} 2009)u^3 + (\log_{10} 3)u^2 + (\log_{10} 28)u - \log_{10}168756 = 0$

Here, we have a cubic function. Since we only want to know the number of solutions to the equation, we can use the cubic function discriminant.

$\Delta = b^2c^2 - 4ac^3 - 4b^3d- 27a^2d^2 + 18 abcd\\ =(\log_{10} 3)^2(\log_{10}28)^2 - 4(\log_{10}2009)(\log_{10}28)^3 - 4(\log_{10}3)^3 (\log_{10}168756)\\ -27(\log_{10}2009)^2(\log_{10}168756)^2 + 18 (\log_{10}2009)(\log_{10} 3)(\log_{10} 28)( \log_{10}168756)\\ \approx -7875.91 < 0$

The value of the discriminant is negative, therefore the equation has only 1 real root and 2 complex roots.

We assume that the question asks for distinct real values of $u$ that satisfies the equation.

Therefore, only $\boxed{1}$ value of $u$ satisfies the equation.